Fields: Prove $1+1=0$

Question

Suppose F has four elements Prove that $1+1=0$

I know that for $F=\{0,1\}$, $1+1 =0$ because if $1+1 =1$, then $1=0$, which is not true, by the definition of field

Answer 1

Any field $F$ is an abelian group under its addition operation. There are two nonisomorphic groups of order $4$, namely $\mathbb{Z}_2 \times \mathbb{Z}_2$ or $\mathbb{Z}_4$.

If the former is the case, then we are done as every nonidentity element has order $2$. Can the latter be true? If it is, then either $a$ or $b$ must have order $2$. In other words, either $a+a = 0$ or $b+b = 0$. From this, we can use the distributive property to write either $a(1+1) = 0$ or $b(1+1) = 0$.

To finish up, you'll want to prove that if $x$ and $y$ are elements of any field, then $xy = 0 \iff x = 0$ or $y=0$.

Answer 2

the characteristic of a finite ring divides the order of the ring.

The characteristic of a finite domain is a prime.